Answer to Question #173180 in Calculus for Hazer

Question

A. Find the area underneath the given curve.

1.) y= -2x² + 8

from x=0 to x=1

2.) y= -x² + 7

from x=0 to x=1

B. Find the area enclosed by the given curve, the x-axis, and the given lines.

1.) y= -1/3x² + 10

from x=1 and x=3

2.) y= x³ - 8

from x= -1 to x =2

C. Find the area bounded by the given curve and line.

1.) y= x² + 3 and y= 7

2.) y= 2x² and y= 4x + 6

3.) y= -x² + 4x and y= x²

4.) y= x³ -6x² + 8x and y= x² -4x

D. Find the area bounded by y=2; y=√x+2; and y=2 - x.

Answer 1

A)

1)

$\intop_0^1(2X^2+8)\\ =8.66$

2)

$\int_0^1-x^2+7\\={20\over3}=6.667$

B)

1)

$y={-1\over 3}x^2+10 \\x=1\\x=3 \\ \therefore\int_1^3({-1\over 3}x^2 +10) \\\boxed{=17.11}$

2)

$y_{1}=x^{3}-8\\ x=1\\ x=2 \\ \therefore \int_1^2(x^3-8x) \\\boxed{=-4.5}$

C]

1)

$y_{1}=x^{2}+3 \\y=7 \\\therefore\int_{-2}^{2}\left(7-x^{2}+3\right)dx\\\boxed{=34.667}$

2)

$y_1=2x^2\\ y_2=4x+6\\ \int_{-1}^{3}\left(y_{1}-y_{2}\right)dx \\= -21.33$

3)

$y_{1}=x^{2}\\y_{2}=-x^{2}+4x\\\int_{0}^{2}\left(y_{2}-y_{1}\right)dx\\=2.66$

4)

$y_{1}=x^{3}-6x^{2}+8x\\y_{2}=x^{2}-4x\\\int_{0}^{3}\left(y_{1}-y_{2}\right)dx+\int_{3}^{4}\left(y_{2}-y_{1}\right)dx\\\boxed{=11.833}$

D}

$y=2\\ y_{2}=\sqrt{x}+2\\y_{3}=2-x$

these curves has no area enclosed so area=0